Integrand size = 31, antiderivative size = 135 \[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx=\frac {1}{8} a^2 (7 A+8 B) x+\frac {a^2 (4 A+5 B) \sin (c+d x)}{3 d}+\frac {a^2 (7 A+8 B) \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a^2 (5 A+4 B) \cos ^2(c+d x) \sin (c+d x)}{12 d}+\frac {A \cos ^3(c+d x) \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{4 d} \] Output:
1/8*a^2*(7*A+8*B)*x+1/3*a^2*(4*A+5*B)*sin(d*x+c)/d+1/8*a^2*(7*A+8*B)*cos(d *x+c)*sin(d*x+c)/d+1/12*a^2*(5*A+4*B)*cos(d*x+c)^2*sin(d*x+c)/d+1/4*A*cos( d*x+c)^3*(a^2+a^2*sec(d*x+c))*sin(d*x+c)/d
Time = 0.27 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.64 \[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx=\frac {a^2 (84 A c+84 A d x+96 B d x+24 (6 A+7 B) \sin (c+d x)+48 (A+B) \sin (2 (c+d x))+16 A \sin (3 (c+d x))+8 B \sin (3 (c+d x))+3 A \sin (4 (c+d x)))}{96 d} \] Input:
Integrate[Cos[c + d*x]^4*(a + a*Sec[c + d*x])^2*(A + B*Sec[c + d*x]),x]
Output:
(a^2*(84*A*c + 84*A*d*x + 96*B*d*x + 24*(6*A + 7*B)*Sin[c + d*x] + 48*(A + B)*Sin[2*(c + d*x)] + 16*A*Sin[3*(c + d*x)] + 8*B*Sin[3*(c + d*x)] + 3*A* Sin[4*(c + d*x)]))/(96*d)
Time = 0.75 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.01, number of steps used = 11, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.355, Rules used = {3042, 4505, 3042, 4484, 25, 3042, 4274, 3042, 3115, 24, 3117}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \cos ^4(c+d x) (a \sec (c+d x)+a)^2 (A+B \sec (c+d x)) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^2 \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^4}dx\) |
| \(\Big \downarrow \) 4505 |
| \(\displaystyle \frac {1}{4} \int \cos ^3(c+d x) (\sec (c+d x) a+a) (a (5 A+4 B)+2 a (A+2 B) \sec (c+d x))dx+\frac {A \sin (c+d x) \cos ^3(c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{4} \int \frac {\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right ) \left (a (5 A+4 B)+2 a (A+2 B) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^3}dx+\frac {A \sin (c+d x) \cos ^3(c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{4 d}\) |
| \(\Big \downarrow \) 4484 |
| \(\displaystyle \frac {1}{4} \left (\frac {a^2 (5 A+4 B) \sin (c+d x) \cos ^2(c+d x)}{3 d}-\frac {1}{3} \int -\cos ^2(c+d x) \left (3 (7 A+8 B) a^2+4 (4 A+5 B) \sec (c+d x) a^2\right )dx\right )+\frac {A \sin (c+d x) \cos ^3(c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{4 d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{3} \int \cos ^2(c+d x) \left (3 (7 A+8 B) a^2+4 (4 A+5 B) \sec (c+d x) a^2\right )dx+\frac {a^2 (5 A+4 B) \sin (c+d x) \cos ^2(c+d x)}{3 d}\right )+\frac {A \sin (c+d x) \cos ^3(c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{3} \int \frac {3 (7 A+8 B) a^2+4 (4 A+5 B) \csc \left (c+d x+\frac {\pi }{2}\right ) a^2}{\csc \left (c+d x+\frac {\pi }{2}\right )^2}dx+\frac {a^2 (5 A+4 B) \sin (c+d x) \cos ^2(c+d x)}{3 d}\right )+\frac {A \sin (c+d x) \cos ^3(c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{4 d}\) |
| \(\Big \downarrow \) 4274 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{3} \left (3 a^2 (7 A+8 B) \int \cos ^2(c+d x)dx+4 a^2 (4 A+5 B) \int \cos (c+d x)dx\right )+\frac {a^2 (5 A+4 B) \sin (c+d x) \cos ^2(c+d x)}{3 d}\right )+\frac {A \sin (c+d x) \cos ^3(c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{3} \left (4 a^2 (4 A+5 B) \int \sin \left (c+d x+\frac {\pi }{2}\right )dx+3 a^2 (7 A+8 B) \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx\right )+\frac {a^2 (5 A+4 B) \sin (c+d x) \cos ^2(c+d x)}{3 d}\right )+\frac {A \sin (c+d x) \cos ^3(c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{4 d}\) |
| \(\Big \downarrow \) 3115 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{3} \left (4 a^2 (4 A+5 B) \int \sin \left (c+d x+\frac {\pi }{2}\right )dx+3 a^2 (7 A+8 B) \left (\frac {\int 1dx}{2}+\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )\right )+\frac {a^2 (5 A+4 B) \sin (c+d x) \cos ^2(c+d x)}{3 d}\right )+\frac {A \sin (c+d x) \cos ^3(c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{4 d}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{3} \left (4 a^2 (4 A+5 B) \int \sin \left (c+d x+\frac {\pi }{2}\right )dx+3 a^2 (7 A+8 B) \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )\right )+\frac {a^2 (5 A+4 B) \sin (c+d x) \cos ^2(c+d x)}{3 d}\right )+\frac {A \sin (c+d x) \cos ^3(c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{4 d}\) |
| \(\Big \downarrow \) 3117 |
| \(\displaystyle \frac {1}{4} \left (\frac {a^2 (5 A+4 B) \sin (c+d x) \cos ^2(c+d x)}{3 d}+\frac {1}{3} \left (\frac {4 a^2 (4 A+5 B) \sin (c+d x)}{d}+3 a^2 (7 A+8 B) \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )\right )\right )+\frac {A \sin (c+d x) \cos ^3(c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{4 d}\) |
Input:
Int[Cos[c + d*x]^4*(a + a*Sec[c + d*x])^2*(A + B*Sec[c + d*x]),x]
Output:
(A*Cos[c + d*x]^3*(a^2 + a^2*Sec[c + d*x])*Sin[c + d*x])/(4*d) + ((a^2*(5* A + 4*B)*Cos[c + d*x]^2*Sin[c + d*x])/(3*d) + ((4*a^2*(4*A + 5*B)*Sin[c + d*x])/d + 3*a^2*(7*A + 8*B)*(x/2 + (Cos[c + d*x]*Sin[c + d*x])/(2*d)))/3)/ 4
Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* x]*((b*Sin[c + d*x])^(n - 1)/(d*n)), x] + Simp[b^2*((n - 1)/n) Int[(b*Sin [c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[ 2*n]
Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[a Int[(d*Csc[e + f*x])^n, x], x] + Simp[b/d In t[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( a_))*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[A*a*Cot[e + f*x]*((d*Csc[e + f*x])^n/(f*n)), x] + Simp[1/(d*n) Int[(d*Csc[e + f*x])^( n + 1)*Simp[n*(B*a + A*b) + (B*b*n + A*a*(n + 1))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && LeQ[n, -1]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[a*A*Cot [e + f*x]*(a + b*Csc[e + f*x])^(m - 1)*((d*Csc[e + f*x])^n/(f*n)), x] - Sim p[b/(a*d*n) Int[(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^(n + 1)*Sim p[a*A*(m - n - 1) - b*B*n - (a*B*n + A*b*(m + n))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0 ] && GtQ[m, 1/2] && LtQ[n, -1]
Time = 0.64 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.56
| method | result | size |
| parallelrisch | \(\frac {a^{2} \left (16 \left (A +B \right ) \sin \left (2 d x +2 c \right )+\frac {8 \left (2 A +B \right ) \sin \left (3 d x +3 c \right )}{3}+A \sin \left (4 d x +4 c \right )+8 \left (6 A +7 B \right ) \sin \left (d x +c \right )+28 \left (A +\frac {8 B}{7}\right ) d x \right )}{32 d}\) | \(75\) |
| risch | \(\frac {7 a^{2} A x}{8}+a^{2} x B +\frac {3 a^{2} A \sin \left (d x +c \right )}{2 d}+\frac {7 \sin \left (d x +c \right ) B \,a^{2}}{4 d}+\frac {A \,a^{2} \sin \left (4 d x +4 c \right )}{32 d}+\frac {A \,a^{2} \sin \left (3 d x +3 c \right )}{6 d}+\frac {\sin \left (3 d x +3 c \right ) B \,a^{2}}{12 d}+\frac {A \,a^{2} \sin \left (2 d x +2 c \right )}{2 d}+\frac {\sin \left (2 d x +2 c \right ) B \,a^{2}}{2 d}\) | \(135\) |
| derivativedivides | \(\frac {A \,a^{2} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+B \,a^{2} \sin \left (d x +c \right )+\frac {2 A \,a^{2} \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+2 B \,a^{2} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+A \,a^{2} \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {B \,a^{2} \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}}{d}\) | \(154\) |
| default | \(\frac {A \,a^{2} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+B \,a^{2} \sin \left (d x +c \right )+\frac {2 A \,a^{2} \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+2 B \,a^{2} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+A \,a^{2} \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {B \,a^{2} \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}}{d}\) | \(154\) |
| norman | \(\frac {\frac {a^{2} \left (7 A +8 B \right ) x}{8}-\frac {7 a^{2} \left (A +8 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{6 d}+\frac {5 a^{2} \left (7 A +8 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{12 d}+\frac {a^{2} \left (7 A +8 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{4 d}+\frac {a^{2} \left (7 A +8 B \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{4}-\frac {a^{2} \left (7 A +8 B \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{8}-\frac {a^{2} \left (7 A +8 B \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{2}-\frac {a^{2} \left (7 A +8 B \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{8}+\frac {a^{2} \left (7 A +8 B \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{4}+\frac {a^{2} \left (7 A +8 B \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}}{8}-\frac {a^{2} \left (25 A +8 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{6 d}+\frac {a^{2} \left (25 A +24 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}-\frac {a^{2} \left (67 A +8 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{12 d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{4} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{2}}\) | \(342\) |
Input:
int(cos(d*x+c)^4*(a+a*sec(d*x+c))^2*(A+B*sec(d*x+c)),x,method=_RETURNVERBO SE)
Output:
1/32*a^2*(16*(A+B)*sin(2*d*x+2*c)+8/3*(2*A+B)*sin(3*d*x+3*c)+A*sin(4*d*x+4 *c)+8*(6*A+7*B)*sin(d*x+c)+28*(A+8/7*B)*d*x)/d
Time = 0.13 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.67 \[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx=\frac {3 \, {\left (7 \, A + 8 \, B\right )} a^{2} d x + {\left (6 \, A a^{2} \cos \left (d x + c\right )^{3} + 8 \, {\left (2 \, A + B\right )} a^{2} \cos \left (d x + c\right )^{2} + 3 \, {\left (7 \, A + 8 \, B\right )} a^{2} \cos \left (d x + c\right ) + 8 \, {\left (4 \, A + 5 \, B\right )} a^{2}\right )} \sin \left (d x + c\right )}{24 \, d} \] Input:
integrate(cos(d*x+c)^4*(a+a*sec(d*x+c))^2*(A+B*sec(d*x+c)),x, algorithm="f ricas")
Output:
1/24*(3*(7*A + 8*B)*a^2*d*x + (6*A*a^2*cos(d*x + c)^3 + 8*(2*A + B)*a^2*co s(d*x + c)^2 + 3*(7*A + 8*B)*a^2*cos(d*x + c) + 8*(4*A + 5*B)*a^2)*sin(d*x + c))/d
Timed out. \[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx=\text {Timed out} \] Input:
integrate(cos(d*x+c)**4*(a+a*sec(d*x+c))**2*(A+B*sec(d*x+c)),x)
Output:
Timed out
Time = 0.03 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.07 \[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx=-\frac {64 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A a^{2} - 3 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{2} - 24 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{2} + 32 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B a^{2} - 48 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{2} - 96 \, B a^{2} \sin \left (d x + c\right )}{96 \, d} \] Input:
integrate(cos(d*x+c)^4*(a+a*sec(d*x+c))^2*(A+B*sec(d*x+c)),x, algorithm="m axima")
Output:
-1/96*(64*(sin(d*x + c)^3 - 3*sin(d*x + c))*A*a^2 - 3*(12*d*x + 12*c + sin (4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*A*a^2 - 24*(2*d*x + 2*c + sin(2*d*x + 2*c))*A*a^2 + 32*(sin(d*x + c)^3 - 3*sin(d*x + c))*B*a^2 - 48*(2*d*x + 2*c + sin(2*d*x + 2*c))*B*a^2 - 96*B*a^2*sin(d*x + c))/d
Time = 0.17 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.30 \[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx=\frac {3 \, {\left (7 \, A a^{2} + 8 \, B a^{2}\right )} {\left (d x + c\right )} + \frac {2 \, {\left (21 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 24 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 77 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 88 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 83 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 136 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 75 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 72 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{4}}}{24 \, d} \] Input:
integrate(cos(d*x+c)^4*(a+a*sec(d*x+c))^2*(A+B*sec(d*x+c)),x, algorithm="g iac")
Output:
1/24*(3*(7*A*a^2 + 8*B*a^2)*(d*x + c) + 2*(21*A*a^2*tan(1/2*d*x + 1/2*c)^7 + 24*B*a^2*tan(1/2*d*x + 1/2*c)^7 + 77*A*a^2*tan(1/2*d*x + 1/2*c)^5 + 88* B*a^2*tan(1/2*d*x + 1/2*c)^5 + 83*A*a^2*tan(1/2*d*x + 1/2*c)^3 + 136*B*a^2 *tan(1/2*d*x + 1/2*c)^3 + 75*A*a^2*tan(1/2*d*x + 1/2*c) + 72*B*a^2*tan(1/2 *d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^4)/d
Time = 11.78 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.99 \[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx=\frac {7\,A\,a^2\,x}{8}+B\,a^2\,x+\frac {3\,A\,a^2\,\sin \left (c+d\,x\right )}{2\,d}+\frac {7\,B\,a^2\,\sin \left (c+d\,x\right )}{4\,d}+\frac {A\,a^2\,\sin \left (2\,c+2\,d\,x\right )}{2\,d}+\frac {A\,a^2\,\sin \left (3\,c+3\,d\,x\right )}{6\,d}+\frac {A\,a^2\,\sin \left (4\,c+4\,d\,x\right )}{32\,d}+\frac {B\,a^2\,\sin \left (2\,c+2\,d\,x\right )}{2\,d}+\frac {B\,a^2\,\sin \left (3\,c+3\,d\,x\right )}{12\,d} \] Input:
int(cos(c + d*x)^4*(A + B/cos(c + d*x))*(a + a/cos(c + d*x))^2,x)
Output:
(7*A*a^2*x)/8 + B*a^2*x + (3*A*a^2*sin(c + d*x))/(2*d) + (7*B*a^2*sin(c + d*x))/(4*d) + (A*a^2*sin(2*c + 2*d*x))/(2*d) + (A*a^2*sin(3*c + 3*d*x))/(6 *d) + (A*a^2*sin(4*c + 4*d*x))/(32*d) + (B*a^2*sin(2*c + 2*d*x))/(2*d) + ( B*a^2*sin(3*c + 3*d*x))/(12*d)
Time = 0.15 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.79 \[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx=\frac {a^{2} \left (-6 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} a +27 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a +24 \cos \left (d x +c \right ) \sin \left (d x +c \right ) b -16 \sin \left (d x +c \right )^{3} a -8 \sin \left (d x +c \right )^{3} b +48 \sin \left (d x +c \right ) a +48 \sin \left (d x +c \right ) b +21 a d x +24 b d x \right )}{24 d} \] Input:
int(cos(d*x+c)^4*(a+a*sec(d*x+c))^2*(A+B*sec(d*x+c)),x)
Output:
(a**2*( - 6*cos(c + d*x)*sin(c + d*x)**3*a + 27*cos(c + d*x)*sin(c + d*x)* a + 24*cos(c + d*x)*sin(c + d*x)*b - 16*sin(c + d*x)**3*a - 8*sin(c + d*x) **3*b + 48*sin(c + d*x)*a + 48*sin(c + d*x)*b + 21*a*d*x + 24*b*d*x))/(24* d)